Quadratic functions

A quadratic function is a function that can be written in the form f(x) = ax² + bx + c, where a, b, c are real numbers and a ≠ 0. Just as a linear function is always described by a straight line, a quadratic function is always described by a parabola.

An example of a quadratic function

An example of a simple quadratic function is f(x) = x² + 3x − 7. The graph of this function looks like this:

We could write a quadratic function in general form as ax² + bx + c. The term ax² is called the quadratic term, and every quadratic function must have it. The next term, bx, is called the linear term and it doesn’t have to appear in a quadratic function – it can be zero. For example, the function f(x) = 5x² + 7 is still a quadratic function. The last term, c, is called the constant term and it isn’t compulsory either. For the function f(x) = x² + 3x − 7 we would therefore have

$$a=1,b=3,c=-7$$

Why is a equal to one? Because writing x² really means 1 · x², so a = 1.

Properties of a quadratic function

Let’s start with the domain. The domain of a quadratic function is ℝ, the set of real numbers. The range depends on the particular function, but it always goes off to (plus or minus) infinity on one side. A quadratic function is always increasing on one side of its turning point and decreasing on the other. If the linear term is zero (b = 0), the quadratic function is even. A quadratic function is never one-to-one.

Bounded above or below

A quadratic function is always bounded either above or below. It depends only on the parameter a. If the parameter a is positive, the graph “opens upwards”, it looks like the letter “U”, and so the graph is bounded below. An example is the function f(x) = 2x² with this graph:

We can see that all the function values (all the red points) are above the number −1, so the function is bounded below. If we had the function f(x) = −2x²:

then all the function values would be below the number 1, so the function would be bounded above.

Convexity and concavity

Convexity and concavity again depend on the parameter a. A quadratic function is convex if it has the shape of the letter “U” and concave if it has the shape of an upside-down “U”. So if a > 0 the graph is convex, and if a < 0 the graph is concave.

The parameter a also affects whether the graph is narrow or wide. The closer its value is to zero, the wider the graph, and vice versa.

Intercepts with the x-axis and y-axis

If we want to work out where a function meets the x-axis or y-axis, we just set up a simple equation and solve it. We only need to realise that we can write the function as y = ax² + bx + c, which means we find the y-coordinate at x = 2 by substituting 2 for every x.

For example, for the function y = x² − 4x + 3 at x = 2 we get the y-coordinate 2² − 4 · 2 + 3 = −1. So the graph certainly passes through the point (2, −1). If we want to work out the x-intercepts, we are interested in the x-value when y = 0. A point on the x-axis always has y-coordinate 0 – look at the following graph and it will be clear.

So in y = x² − 4x + 3 we just substitute zero for y and solve the equation. The resulting roots are the x-coordinates of the intercepts.

$$x^2-4x+3=0$$

This is a quadratic equation, which we can solve by standard methods. For example, we can factorise it as (x − 1) · (x − 3), from which we find that the roots are x₁ = 1 and x₂ = 3.

To get the y-intercept we proceed in the same way. We take y = x² − 4x + 3 and this time substitute zero for x. We are really asking what the value of the function is at x = 0:

$$0^2-4\cdot 0+3=3$$

So the x-intercepts are (1, 0) and (3, 0), and the y-intercept is (0, 3).

How to find the coordinates of the turning point

For a quadratic function it is also very important to find its turning point (the vertex). So that you don’t have to remember a rather complicated formula, we’ll show you another way, which uses the method of completing the square. The turning point of a quadratic function is the point where the function has a minimum or maximum. The function f(x) = x² − 4x + 3, whose graph looks like this:

has its turning point at (2, −1).

The standard way to write a quadratic function is f(x) = ax² + bx + c. For a = 1, we convert this function into the completed-square form g(x) = (x + m)² + n, where (−m, n) is the turning point of the quadratic function. For our function we want to get (x − 2)² − 1. How do we get this form?

First we write down just the bracket: (x + m), and for m we substitute half the value of the parameter b of the quadratic function. The parameter b is equal to minus four, half of it is minus two, so we get this expression: (x − 2)².

Now comes the second step: we have to adjust for the extra terms. If you expanded this bracket, you wouldn’t get the right result – we wouldn’t get the original function. From the bracket we still have to subtract the square of m and add the parameter c of the original function.

So specifically, we subtract (−2)² from the bracket and add three (the parameter c):

$$(x-2{)}^2-4+3$$

After simplifying we get

$$(x-2{)}^2-1,$$

which is the final result. Because the function is in the form (x + m)² + n, where m = −2, n = −1, we know that the turning point has coordinates (−m, n), that is, (2, −1).

We should still check that we worked it out correctly. All we did was convert the quadratic function x² − 4x + 3 into the form (x − 2)² − 1. This new form should describe the same function, so if we expand the bracket, we should get the original form of the function. Let’s try expanding (x − 2)² − 1.

$$\begin{array}{rcl}(x-2{)}^2-1& =& (x-2)\cdot (x-2)-1\\ & =& x^2-4x+4-1\\ & =& x^2-4x+3\end{array}$$

We can see that we got the same function, so our working is correct.